First slide

Capacitance

Question

A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

Moderate

A

0%
B

20%
C

75%
D

80%

Solution

Initial energy stored in capacitor 2μF
$U_{i} = \frac{1}{2} 2 {(V)}^{2} = V^{2}$
Final voltage after switch 2 is ON
$V_{f} = \frac{C_{1} V_{1}}{C_{1} + C_{2}} = \frac{2 V}{10} = 0.2 V$
Final energy in both the capacitors
$U_{f} = \frac{1}{2} (C_{1} + C_{2}) {V_{f}}^{2} = \frac{1}{2} 10 {(\frac{2 V}{10})}^{2} = 0.2 V^{2}$
So, energy dissipated = $\frac{V^{2} - 0.2 V^{2}}{V^{2}} x 100 = 80 %$

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